. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i

Answer:

Step-by-step explanation:

Hello!

The monitoring system alerts the driver when the vehicle's tire pressure drops to 28% below the set target pressure.

Let X be: the target tire pressure for a particular vehicle (measured in pounds per square inch)

a)

X= 28 psi

When the monitoring device alerts at a pressure of 28% below the designated target: X-0.28X

Initially, calculate 28% of 28 psi.

28*0.28= 7.84

Next, subtract this 28% calculation from the target pressure:

28 - 7.84= 20.16

The TPMS will activate its warning at 20.16 psi.

b)

Assuming X~N(μ;σ²)

μ= 28 psi (as the average indicates accurate targeting, this represents the expected tire pressure)

σ= 3 psi

P(X≤20.16)

The standard normal distribution is available in tables. To convert any random variable X with a normal distribution, one subtracts its mean and divides by the standard deviation.

To compute the desired probabilities, the variable value undergoes transformation to fit the standard normal distribution Z, after which standard normal tables are referenced to find probabilities.

Z= (X-μ)/σ= (20.16-28)/3= -2.61

Now locate the probability corresponding to the Z value using the Z-table. Given the negative result, refer to the left entry; in the first column, find the integer and first decimal for -2.6- and in the first row locate the second decimal for -.-1

The probability link for -2.61 is:

P(Z≤-2.61)= 0.005

c)

The task is to determine the likelihood of encountering a tire at random within the recommended inflation range, expressed as:

P(30≤X≤26)= P(X≤30)-P(X≤26)

Determine both Z values:

Z= (30-28)/3= 0.67

Z= (26-28)/3= -0.67

P(Z≤0.67)-P(Z≤-0.67)= 0.749 - 0.251= 0.498

<pThe calculated probability of a tire being inflated within the suggested range is 0.498.

I hope this information is useful!

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At this given point, the derivative value of the Hessian matrix is



and the second-order partial derivative with respect to yields



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To find the hottest area on the plate, it must be located along the boundary. Let and , so that




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Our minimum has already been identified inside the plate - which you can check to have a lower temperature than at the points noted by - and we identify two maxima at and , both showing a maximum temperature of .

Reverting to Cartesian coordinates, these points match up with .

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